How to calculate the following limit - I'm stuck!

  • Thread starter chemic_23
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  • #1
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Homework Statement



what is the limit of
factoring.jpg
as x approaches 9?

Homework Equations





The Attempt at a Solution



factoringsol.jpg
I'm stuck please help...

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
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I would do it this way:
remember that:
a[tex]^{2}[/tex]-b[tex]^{2}[/tex]=(a+b)(a-b)
so
a-b=[tex]\frac{(a^2-b^2}{(a+b)}[/tex]

so take the numerator and denominator of your problem and treat each as a-b
so:
[tex]\sqrt{x}-3[/tex] = [tex]\frac{x-9}{\sqrt{x}+3}[/tex] (1)

and
[tex]\sqrt{1+\sqrt{x}}-2[/tex] = [tex]\frac{\sqrt{x} - 3}{\sqrt{1+\sqrt{x}}+2}[/tex] (2)

so your limit is now (1)/(2): which gives:

[tex]\frac{x-9}{\sqrt{x}+3}\times\frac{\sqrt{1+\sqrt{x}}+2}{\sqrt{x}-3}[/tex]

Is this clear?
Do you know how to continue from here?
 
  • #3
44
0


continuation...
sol2.jpg


thus, the limit of
giv.jpg
as x approaches 9 is 4. Is this correct?
 
  • #4
Dick
Science Advisor
Homework Helper
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619


Correct.
 

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